How do you solve the given equation $ {{e}^{2x}}-2{{e}^{x}}+1=0 $ ?

Hint: We start solving the problem by first factorizing the given equation by making use of the facts that \[{{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}\], $ a\times a={{a}^{2}} $ . We then make use of the fact that if $ {{a}^{2}}=0 $ , then $ a=0 $ to proceed through the problem. We then make the necessary calculations and make use of the fact that that if $ {{a}^{m}}={{a}^{n}} $ , then $ m=n $ to get the required answer.

Complete step by step answer:
According to the problem, we are asked to solve the given equation $ {{e}^{2x}}-2{{e}^{x}}+1=0 $ .
So, we have $ {{e}^{2x}}-2{{e}^{x}}+1=0 $ .
Let us factorize this equation to find the value of x which satisfies the given equation.
 $ \Rightarrow {{e}^{2x}}-{{e}^{x}}-{{e}^{x}}+1=0 $ .
 $ \Rightarrow {{e}^{2\times x}}-{{e}^{x}}-{{e}^{x}}+1=0 $ ---(1).
From laws of exponents, we know that \[{{a}^{m\times n}}={{\left( {{a}^{m}} \right)}^{n}}={{\left( {{a}^{n}} \right)}^{m}}\]. Let us use this result in equation (1).
 $ \Rightarrow {{\left( {{e}^{x}} \right)}^{2}}-{{e}^{x}}-{{e}^{x}}+1=0 $ .
 $ \Rightarrow {{e}^{x}}\left( {{e}^{x}}-1 \right)-1\left( {{e}^{x}}-1 \right)=0 $ .
 $ \Rightarrow \left( {{e}^{x}}-1 \right)\left( {{e}^{x}}-1 \right)=0 $ ---(2).
We know that $ a\times a={{a}^{2}} $ . Let us use this result in equation (2).
 $ \Rightarrow {{\left( {{e}^{x}}-1 \right)}^{2}}=0 $ ---(3).
We know that if $ {{a}^{2}}=0 $ , then $ a=0 $ . Let us use this result in equation (1).
 $ \Rightarrow {{e}^{x}}-1=0 $ .
 $ \Rightarrow {{e}^{x}}=1 $ ---(4).
From laws of exponents, we know that $ {{e}^{0}}=1 $ . Let us use this result in equation (4).
 $ \Rightarrow {{e}^{x}}={{e}^{0}} $ ---(5).
From laws of exponents, we know that if $ {{a}^{m}}={{a}^{n}} $ , then $ m=n $ . Let us use this result in equation (5).
 $ \Rightarrow x=0 $ .
So, we have found the solution of the given equation $ {{e}^{2x}}-2{{e}^{x}}+1=0 $ as $ x=0 $ .
$ \therefore $ The solution of the given equation $ {{e}^{2x}}-2{{e}^{x}}+1=0 $ is $ x=0 $ .

Note:
 We should perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we get this type of problem, we first try to factorize the given equation and then equate each factor to zero to get the required solution for that equation. We should keep in mind that $ {{e}^{x}}\ge 0 $ while solving the problems involving exponents. Similarly, we can expect problems to find the solution for the equation $ {{e}^{2x}}-1=0 $ .